With finite objectives, I have generally used these on a bellows. By lengthening or shortening the bellows, the magnification can be increased or decreased, respectively, and it is pretty easy to visually understand what's going on.

I just (within the last few weeks) began to use infinity-corrected objectives. I have several options of tube lens to pair them with: with a Canon EF 100mm or 180mm macro lens, or with a Raynox DCR-150 (208mm) or DCR-250 (125mm).

One thing I noticed is that again, I am able to adjust magnification by putting a bellows between the tube lens and camera. I can't imagine that this would differ in any way from doing this with a finite lens without a tube lens... or adding extension tubes to a camera lens... but correct me if I'm wrong.

Now, about the math. To calculate the effective magnification for finite objectives, I've been using a formula that requires the focal length of the objective, its rated magnification, the actual optical tube length, and the rated optical tube length. For finite objectives, you can just look the focal length up in the technical specifications. But for infinite objectives, which are only one part of an optical assembly... what focal length do I use? Do I only use the focal length of the tube lens, or only of the objective, or some mathematical combination of the two?

Another interesting thing I've noticed is that when using my Canon EF 100mm f/2.8L macro or 180mm f/3.5L macro as tube lenses, focusing closer than infinity reduces the magnification yet still allows apparently good image formation. I was a bit surprised by the change in magnification here and welcome discussion on why this occurs.

One thing I noticed is that again, I am able to adjust magnification by putting a bellows between the tube lens and camera. I can't imagine that this would differ in any way from doing this with a finite lens without a tube lens... or adding extension tubes to a camera lens... but correct me if I'm wrong.

This is correct. A good model is that infinite objective + tube lens = finite objective. Moving any objective away from its design point will introduce some aberrations, but those largely depend on details of the lens designs that we never know anyway, so if it matters you have to test.

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But for infinite objectives, which are only one part of an optical assembly... what focal length do I use? Do I only use the focal length of the tube lens, or only of the objective, or some mathematical combination of the two?

Doing it with just math is difficult. The focal length of the combination depends in part on the separation between principal planes of the objective and the tube lens, and the locations of those things fall in the category of "details that we never know anyway". (There's a formula given at Wikipedia, compound lenses, but note that their "d" actually refers to separation between principal planes, not just the length of the air gap as you might think.)

However, it's fairly straightforward to determine the effective focal length by measuring the magnifications at two different known extensions, then calculating as:

EFL = (exten1 - exten2) / (mag1 - mag2)

In words, the effective focal length can be measured as the difference in extensions divided by the difference in magnifications.

Once you know the effective focal length, then you can calculate the magnification at any extension by:

mag3 = mag1 + (exten3-exten1)/EFL

Unless, of course, I've messed up the algebra. Let me know if that's the case...

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Another interesting thing I've noticed is that when using my Canon EF 100mm f/2.8L macro or 180mm f/3.5L macro as tube lenses, focusing closer than infinity reduces the magnification yet still allows apparently good image formation. I was a bit surprised by the change in magnification here and welcome discussion on why this occurs.

This occurs because you have lenses whose focal length gets shorter as you turn the ring to make them focus closer. As an example my particular EF 100 mm f/2.8L starts off at just under 105 mm (measured as described above, using infinity focus on the lens ring), then gradually shortens to about 73 mm at closest focus (using 1:1 on the lens ring). When such a lens is paired with an objective, you end up with a combination of two effects: 1) there is a minor increase in magnification caused by slightly reducing the subject-to-objective distance, but this is more than countered by 2) there is a significant decrease in magnification caused by shortening the focal length of the tube lens.

In addition to altering magnification, there will also be some amount of added aberration due to altering the subject-to-objective distance. The added aberration is not important at say 10X NA 0.25, but can become serious at say 50X NA 0.55. See for example http://www.photomacrography.net/forum/viewtopic.php?p=92144#92144 .

One last thing, note that turning the lens ring can also move the principal planes, and by a significant amount in some designs. Take this into account if you're masochistic enough to try back-calculating "d" based on Wikipedia's formula and other numbers determined as shown above.

On casual glance your math seems correct... and looks to be the same math I was already using! Although I don't bother to do these calculations very often, since if I need to know magnification, I think I'm better off swapping the subject with a stage micrometer and measuring it directly.

I have an excel-based macro calculator that I have always intended to use as a way of figuring out some initial estimate for the number of slices to use in a stack. I've never actually bothered to use it yet, though... I usually end up guessing + overshooting the necessary number of slices.

Last question...

Can we assume that the Raynox lenses have the same front focal length and back focal length? Since they are reversed as use as a tube lens, I am making this assumption in order to determine the necessary optical extension to put the tube lens at infinity focus.

Can we assume that the Raynox lenses have the same front focal length and back focal length?

I'm not sure I understand the question.

Any lens that has the same medium on both sides will have the same front and rear focal length, where those terms have their standard optics definitions that refer to the size of image you'll get if you focus on an object at infinity, or equivalently if you measure the focal length using the magnification-versus-extension test.

On the other hand, it's very unlikely that the front and rear focal points will lie at the same distance away from the lens mount. So, to get infinity focus in the normal versus reversed orientations, you'll almost certainly need different extensions. This is another of those situations where you're better off to measure, not calculate.

It's the distance between the front/rear focal points and the surface of the front or rear lens element.

But yes, measurement is probably best. As far as I can tell the easiest thing to do would be to stick the reversed Raynox on a bellows and judge the extension for infinity focus by changing extension until an image of the horizon is formed. I just did this with the DCR-150, it only takes a moment. Looks to be roughly 211mm of optical tube length extension on an EF mount camera, including the flange distance (166mm of bellows+adapter ring extension). Just measured using a standard ruler, though.

bothers me - if the object is at infinity the image size would be zero, no?

If you mean that anything infinitely far away will be seen as an infinitesimal point, I agree. I'd like to imagine someone's inner monologue long ago:

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I need a word to describe the imaging of an object that is far away... how about "far"? Na, too ambiguous. Oh, look, a thesaurus!

I also like how many lenses can be racked beyond the infinity marks. I wonder what they expect to see out there.

To illustrate how confusing some of these terms are to people who aren't optical engineers, I just now realized that my original question is silly because I was confusing "focal length" and "back focal length", which are quite different because "focal length" accounts for optical design and lens thickness while "back focal length" does not.

It's the distance between the front/rear focal points and the surface of the front or rear lens element.

I stand corrected on the terminology. Indeed it seems that "front focal length" and "back focal length" are defined in terms of the surfaces of lens elements. So now I have to agree with ChrisR that these are silly names, best avoided. In practical lens designs they can be wildly different from the effective focal length. For example my Canon 55-200 mm telephoto has a rear element that is no more then 100 mm away from sensor even when the lens is set to 200 mm focal length. If I stick a teleconverter on my camera, then the rear element of the optics will consistently be about 40 mm away from my sensor, regardless of the effective focal length of the optics. It strikes me as an invitation to confusion to have the phrases "XX focal length" mean such wildly different things depending on what XX happens to be.

ChrisR wrote:

While I'm here interfering though,

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standard optics definitions that refer to the size of image you'll get if you focus on an object at infinity

bothers me - if the object is at infinity the image size would be zero, no?

Not if the object is infinitely large.

By way of explanation... A lens that forms a real image and is "focused at infinity" is in the business of accepting bundles of parallel rays where there is some angular separation between bundles, and turning those into cones of focused rays where there is some linear separation between focused points. Given a fixed angular separation, the focal length of the lens corresponds to the linear separation between the focused points. I'm using "size of the image" as shorthand for that linear separation.

It's certainly true that the magnification would be zero, however. And that's why total extension = focal length * (magnification+1) is true even at infinity focus. (Carefully defining "total extension" as the distance between the rear principal point and the focal plane, of course...)